A cylindrical rod is reformed to half of its original length keeping volume constant. If its resistance before this change were R, then the resistance after reformation of rod will be

Options

A. R
B. R/4
C. 3R/4
D. R/2

Explanation

To solve the problem, we need to analyze the relationship between the resistance of the cylindrical rod before and after it is reformed while keeping the volume constant.

### Step-by-Step Solution:

1. **Understand the Initial Resistance**:
The resistance \( R \) of a cylindrical rod is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( \rho \) is the resistivity of the material, \( L \) is the length of the rod, and \( A \) is the cross-sectional area of the rod. For a cylinder, the area \( A \) can be expressed as:
\[ A = \pi r^2 \]
Thus, the resistance can also be written as:
\[ R = \frac{\rho L}{\pi r^2} \]

2. **Identify the Changes**:
The rod is reformed to half its original length. Let the original length be \( L_1 \) and the new length be \( L_2 = \frac{L_1}{2} \). The volume of the rod remains constant, which means:
\[ V = A_1 L_1 = A_2 L_2 \]
where \( A_1 \) and \( A_2 \) are the cross-sectional areas before and after reformation, respectively.

3. **Set Up the Volume Equation**:
Since the volume is constant:
\[ \pi r_1^2 L_1 = \pi r_2^2 L_2 \]
Simplifying this gives:
\[ r_1^2 L_1 = r_2^2 \left(\frac{L_1}{2}\right) \]
Therefore:
\[ r_1^2 = \frac{r_2^2}{2} \]
This implies:
\[ \frac{r_1^2}{r_2^2} = \frac{1}{2} \quad \Rightarrow \quad \frac{r_2^2}{r_1^2} = 2 \]

4. **Calculate the New Resistance**:
The new resistance \( R_2 \) after reformation is:
\[ R_2 = \frac{\rho L_2}{A_2} = \frac{\rho \left(\frac{L_1}{2}\right)}{\pi r_2^2} \]
Substituting \( A_2 = \pi r_2^2 \) and using \( r_2^2 = 2r_1^2 \):
\[ R_2 = \frac{\rho \left(\frac{L_1}{2}\right)}{\pi (2r_1^2)} = \frac{\rho L_1}{4\pi r_1^2} \]
Now, we can express this in terms of the original resistance \( R \):
\[ R_2 = \frac{1}{4} \left(\frac{\rho L_1}{\pi r_1^2}\right) = \frac{R}{4} \]

5. **Conclusion**:
Therefore, the resistance after the reformation of the rod is:
\[ R_2 = \frac{R}{4} \]

### Final Answer:
The resistance after reformation of the rod will be \( \frac{R}{4} \).